When 80 volt d.c. is applied across a solenoid, a current of 0.8 A flows in it. When 80 volt a.c. is applied…
When 80 volt d.c. is applied across a solenoid, a current of 0.8 A flows in it. When 80 volt a.c. is applied across the same solenoid, the current becomes 0.4 A . If the frequency of a.c. source is 50 Hz , the impedance and inductance of the solenoid is nearly
$200 \Omega, 0.55 \mathrm{H}$
$100 \Omega, 0.8 \mathrm{H}$
$300 \Omega, 1.2 \mathrm{H}$
$200 \Omega, 1.5 \mathrm{H}$
Solution
When 80 Vdc is applied,
$\begin{aligned}
Z & =\sqrt{\mathrm{R}^2+(\omega \mathrm{L})^2}=\mathrm{R} \quad \ldots(\because \omega=0 \text { in dc circuit }) \\
\therefore \quad \mathrm{Z} & =\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{80}{0.8}=100 \Omega \quad \ldots \text { (i) }
\end{aligned}$ When 80 V ac is applied,
$\begin{aligned}
& Z=\frac{V}{I}=\frac{100}{0.5}=200 \Omega ...(ii)\\
& Z=\sqrt{R^2+(\omega L)^2} \\
& \Rightarrow(2 \pi \mathrm{fL})^2=Z^2-R^2 \\
& L=\sqrt{\frac{200^2-100^2}{(2 \pi \times 50)^2}}=0.55 \mathrm{H}
\end{aligned}$
...[From(i).and (ii)]